B. Flag of Berland
1 second
256 megabytes
standard input
standard output
The flag of Berland is such rectangular field n × m that satisfies following conditions:
- Flag consists of three colors which correspond to letters 'R', 'G' and 'B'.
- Flag consists of three equal in width and height stripes, parralel to each other and to sides of the flag. Each stripe has exactly one color.
- Each color should be used in exactly one stripe.
You are given a field n × m, consisting of characters 'R', 'G' and 'B'. Output "YES" (without quotes) if this field corresponds to correct flag of Berland. Otherwise, print "NO" (without quotes).
The first line contains two integer numbers n and m (1 ≤ n, m ≤ 100) — the sizes of the field.
Each of the following n lines consisting of m characters 'R', 'G' and 'B' — the description of the field.
Print "YES" (without quotes) if the given field corresponds to correct flag of Berland . Otherwise, print "NO" (without quotes).
6 5
RRRRR
RRRRR
BBBBB
BBBBB
GGGGG
GGGGG
YES
4 3
BRG
BRG
BRG
BRG
YES
6 7
RRRGGGG
RRRGGGG
RRRGGGG
RRRBBBB
RRRBBBB
RRRBBBB
NO
4 4
RRRR
RRRR
BBBB
GGGG
NO
The field in the third example doesn't have three parralel stripes.
Rows of the field in the fourth example are parralel to each other and to borders. But they have different heights — 2, 1 and 1.
这题还是不错的,题意是如果 可以分割成3条,R G B各一条,那就输出YES,每条可以包含多行,但是每条的行数必须相等
打了一大堆补丁,最后过了
n,m = map(int,raw_input().split())
mark = 1
s = []
for i in range(n):
tmp = raw_input();
s.append(tmp);
for c in range(m):
if c + 1 < m and tmp[c] != tmp[c + 1]:
mark = 0;
r = 0
g = 0
b = 0
for a in s:
for c in a:
if c =='R':
r = r + 1
if c == 'G':
g = g + 1
if c == 'B':
b = b + 1
if mark == 1:
num = 1
w = []
for i in range(n):
if i + 1 < n and s[i][0] == s[i + 1][0]:
num = num + 1;
else:
w.append(int(num))
num = 1;
for i in range(len(w)):
if i + 1 < len(w) and w[i] != w[i + 1]:
mark = 2;
if mark == 1 and len(w) == 3 and r == g and g == b:
print "YES"
else:
print "NO"
else :
for y in range(m):
for x in range(n):
if x + 1 < n and s[x][y] != s[x + 1][y]:
mark = 2;
break;
if mark == 2:
print "NO"
else :
num = 1
w = []
for i in range(m):
if i + 1 < m and s[0][i] == s[0][i + 1]:
num = num + 1
else:
w.append(int(num));
num = 1;
for i in range(len(w)):
if i + 1 < len(w) and w[i] != w[i + 1]:
mark = 2;
if mark == 0 and len(w) == 3 and r == g and g ==b:
print "YES"
else:
print "NO"
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